Problem: The following line passes through point $(1, 9)$ : $y = -\dfrac{13}{3} x + b$ What is the value of the $y$ -intercept $b$ ?
Answer: Substituting $(1, 9)$ into the equation gives: $9 = -\dfrac{13}{3} \cdot 1 + b$ $9 = -\dfrac{13}{3} + b$ $b = 9 + \dfrac{13}{3}$ $b = \dfrac{40}{3}$ Plugging in $\dfrac{40}{3}$ for $b$, we get $y = -\dfrac{13}{3} x + \dfrac{40}{3}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${10}$ ${11}$ ${12}$ ${13}$ ${14}$ ${15}$ ${16}$ ${17}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${\llap{-}10}$ ${\llap{-}11}$ ${\llap{-}12}$ ${\llap{-}13}$ ${\llap{-}14}$ ${\llap{-}15}$ ${\llap{-}16}$ ${\llap{-}17}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${10}$ ${11}$ ${12}$ ${13}$ ${14}$ ${15}$ ${16}$ ${17}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${\llap{-}10}$ ${\llap{-}11}$ ${\llap{-}12}$ ${\llap{-}13}$ ${\llap{-}14}$ ${\llap{-}15}$ ${\llap{-}16}$ ${\llap{-}17}$ $(1, 9)$